Performing the operations indicated, we have a3+6a2b-3 a2c+12 ab2 - 12 abc + 3 ac2+8b3 — 12 b2c +6 bc2 - c3. NOTE. A full discussion of the Binomial Theorem for Positive Integral Index is given in Chapter XXXVII. CHAPTER XXI. EVOLUTION. 195. The root of any proposed expression is that quantity which being repeated as a factor the requisite number of times produces the given expression. (Art. 14.) The operation of finding the root is called Evolution: it is the inverse of Involution. 196. By the Rule of Signs we see that (1) any even root of a positive quantity may be either positive or negative; (2) no negative quantity can have an even root; (3) every odd root of a quantity has the same sign as the quantity itself. NOTE. It is especially worthy of notice that every positive quantity has two square roots equal in magnitude, but opposite in sign. In the present chapter, however, we shall confine our attention to the positive root. EVOLUTION OF MONOMIALS. 197. From a consideration of the following examples we will be able to deduce a general rule for extracting any proposed root of a monomial. EXAMPLES. (1) √(a®b1) = a3b2 because (a3b2)2= a6b4. (2) † (− x9) = − x3 because (− x3)3 — — x9. Rule. (1) Find the root of the coefficient by Arithmetic, and prefix the proper sign found by Art. 42. (2) Divide the exponent of every factor of the expression by the index of the proposed root. It will be seen that in the last case we operate separately upon the numerator and the denominator. EXAMPLES XXI. 8. Write the square root of each of the following expressions: Write the cube root of each of the following expressions: 21. 15. 64x6у3z12. 216 x6y9 512 221 Write the value of each of the following expressions: Since the 198. The Square Root of Any Multinomial. square of a + b is a2 + 2 ab + b2, we have to discover a process by which a and b, the terms of the root, can be found when a2+2ab+b2 is given. The first term, a, is the square root of a2. Arrange the terms according to powers of one letter a. The first term is a2, and its square root is a. Set this N down as the first term of the required root. Subtract a3 from the given expression and the remainder is 2 ab + b2 or (2 a + b) × b. Thus, b, the second term of the root, will be the quotient when the remainder is divided by 2 a+b. This divisor consists of two terms: 1. The double of a, the term of the root already found. 2. b, the new term itself. The work may be arranged as follows: a2+2ab+b2 (a + b Ex. 1. Find the square root of 9 x2 - 42 xy + 49 y2. 9 x2-42xy + 49 y2 (3 x — 7 y EXPLANATION. The square root of 9 x2 is 3x, and this is the first term of the root. By doubling this we obtain 6x, which is the first term of the divisor. Divide 42 xy, the first term of the remainder, by 6 x and we get 7y, the new term in the root, which has to be annexed both to the root and divisor. Next multiply the complete divisor by - 7 y and subtract the result from the first remainder. There is now no remainder and the root has been found. The process can be extended so as to find the square root of any multinomial. The first two terms of the root will be obtained as before. When we have brought down the second remainder, the first part of the new divisor is obtained by doubling the terms of the root already found. We then divide the first term of the remainder by the first term of the new divisor, and set down the result as the next term in the root and in the divisor. We next multiply the complete divisor by the last term of the root and subtract the product from the last remainder. If there is now no remainder the root has been found; if there is a remainder we continue the process. Ex. 2. Find the square root of 25 x2a2-12 xa3 + 16 x + 4 aa — 24x3α. Rearrange in descending powers of x. 16 x 24 x3α + 25 x2a2 - 12 xa3 + 4 a1 (4 x2 - 3 xa + 2 a2 EXPLANATION. 16 x2a2 12 xa3 + 4 a1 16 x2a2-12 xa3 + 4 a1 When we have obtained two terms in the root, 4x2 - 3xa, we have a remainder 16 x2a2 12 xa3 + 4 a1. Double the terms of the root already found and place the result, 8 x2-6xa, as the first part of the divisor. Divide 16x2a2, the first term of the remainder, by 8x2, the first term of the divisor; we get +2 a2 which we annex both to the root and divisor. Now multiply the complete divisor by 2 a2 and subtract. There is no remainder and the root is found. EXAMPLES XXI. b. Find the square root of each of the following expressions: 1. x2-10 xy + 25 y2. 2. 4x2-12xy + 9 y2. 3. 81 x2+18 xy + y2. 4. 25x2-30 xy + 9 y2. 5. a -2 a3 + 3 a2 - 2 a + 1. 10. 110x + 27 x2 - 10 x3 + x4. 11. 4x2+9 y2 + 25 z2 + 12 xy - 30 yz - 20 xz. 12. 16x6 16 x7 - 4 x8 — 4 x9 + x1o. 13. x6 - 22 x1 + 34x3 + 121 x2 - 374x + 289. 16. 6ab2c-4 a2bc + a2b2 + 4 a2c2 + 9 b2c2 - 12 abc2. 18. 4x4 +9 y + 13 x2y2 — 6 xy3 — 4 x3y. 19. 1 4x + 10 x2 - 20 x3 + 25 x1 — 24x5+ 16 x6. 20. 6 acx5+ 4 b2x4 + a2x10 + 9 c2 - 12 bcx2 - 4 abx7. 199. When the expression whose root is required contains fractional terms, we may proceed as before, the fractional |